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Chapter 35: Problem 39

An infinite square well extends from \(-L / 2\) to \(L / 2 .\) (a) Find expressions for the normalized wave functions for a particle of mass \(m\) in this well, giving separate expressions for even and odd quantum numbers. (b) Find the corresponding energy levels.

Short Answer

Expert verified
The normalized wavefunctions for a particle of mass m in the well are \( \psi(x) = \sqrt{\frac{2}{L}} \sin(\frac{n \pi x}{L}) \) for odd quantum numbers and \( \psi(x) = \sqrt{\frac{2}{L}} \cos(\frac{n \pi x}{L}) \) for even quantum numbers. The energy levels are \( E = \frac{n^2 \pi^2 \hbar^2}{2mL^2} \) for \( n = 1,2,3... \).

Step by step solution

01

Obtain the Wavefunction

Applying the time-independent Schrödinger equation in a region where the potential energy is zero, a general solution for the wave function can be obtained: \( \psi(x) = A \sin(kx) + B \cos(kx) \), where \( k = \sqrt{2mE / \hbar^2} \). Since the potential is infinity for \( |x| > L/2 \), the wavefunction has to vanish at \( x = ±L/2 \).
02

Obtain the Normalized Wavefunctions for Odd Quantum Numbers

sin(kx) is obtained for the wave function under odd quantum numbers and has half an integral multiple of wavelengths inside the well. Given the conditions: \( \psi(-L/2) = \psi(L/2) = 0 \), the equation: \( k = \frac{n \pi}{L} \), \( n = 1,3,5... \) can be derived. The normalized wave function then is: \( \psi(x) = \sqrt{\frac{2}{L}} \sin(\frac{n \pi x}{L}) \), for odd n.
03

Obtain the Normalized Wavefunctions for Even Quantum Numbers

cos(kx) is obtained for the wave function under even quantum numbers given the conditions: \( \psi(-L/2) = \psi(L/2) \). Therefore, the equation: \( k = \frac{n \pi}{L} \), \( n = 2,4,6... \) can be derived.The normalized wave function then is: \( \psi(x) = \sqrt{\frac{2}{L}} \cos(\frac{n \pi x}{L}) \), for even n.
04

Find the Energy Levels

Substituting \( k = \frac{n \pi}{L} \) into the expression for \( k = \sqrt{2mE / \hbar^2} \), the energy levels for both odd and even quantum numbers can be found: \( E = \frac{n^2 \pi^2 \hbar^2}{2mL^2} \), for \( n = 1,2,3... \)

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Most popular questions from this chapter

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The ground-state energy for an electron in infinite square well \(A\) is equal to the energy of the first excited state for an electron in well B. How do the wells' widths compare?

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